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What is the molarity of a solution dissolve 0.31 grams of HNO3 in 300ml of water?
Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3
What is the molarity of srno32 in a solution resulting from mixing 150.0 ml of 0.0200 m hno3 with 150.0 ml of 0.0100 m sroh2?
You get 300 mL of 0.0100 M Sr(NO3)2 solution
How would you prepare a 300 ml of a 0.1 M Na CL solution from a solid Na CL solution and water?
You prepare a solution by dissolving a known mass of solute into a specific amount of solvent. In solutions, M is the molarity, or moles of solute per liter of solution. For 300 ml of a 0.1 M Na CL solution from a solid Na CL solution and water you need water and sodium chloride.
Please show work How many moles would there be in 300 ml of of a 1.5 M solution?
To find the number of moles, use the formula: moles = Molarity (M) x Volume (L). First, convert 300 ml to liters by dividing by 1000: 300 ml / 1000 = 0.3 L. Then, calculate moles = 1.5 M x 0.3 L = 0.45 moles. Therefore, there would be 0.45 moles in 300 ml of a 1.5 M solution.
How much magnesium chloride should be added to a liter of water to produce a saturated solution?
Saturated solutions of:anhydrous MgCl254.3 g/100 ml (20 °C)72.6 g/100 mL (100 °C)hexahydrate MgCl2(H2O)6157 g/100 mL (20 °C)Note:[A variety of hydrates are known with the formula MgCl2(H2O)x, and each loses water with increasing temperature:below 0°C (as de-icer): x = 12 (-16.4 °C) and x = 8 (-3.4 °C),above 100°C: x = 6 (116.7 °C), x = 4 (181 °C) and x = 2 (ca. 300 °C)]
What is the molarity of a 3000 liters solution containing 300 grams of NaCl?
Molarity = moles of solute/Liters of solution Find moles NaCl 300 grams NaCl (1 mole NaCl/58.44 grams) = 5.13347 moles NaCl Molarity = 5.13347 moles NaCl/3000 Liters = 1.71 X 10^-3 M sodium chloride ----------------------------------------
Calculate molarity of solution containing 16 gm glucose per 300 ml solution?
Molar mass of glucose (C6H12O6) = 180 g/mol. Number of moles of glucose = 16 g / 180 g/mol = 0.089 moles. Volume of solution = 300 ml = 0.3 L. Molarity = number of moles / volume of solution = 0.089 moles / 0.3 L ≈ 0.297 M.
What is the molarity of a solution dissolve 0.31 grams of HNO3 in 300ml of water?
Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3
What is the molarity of srno32 in a solution resulting from mixing 150.0 ml of 0.0200 m hno3 with 150.0 ml of 0.0100 m sroh2?
You get 300 mL of 0.0100 M Sr(NO3)2 solution
How many moles of potassium hydroxide are required to prepare 300 mL of 0.250 M solution?
To calculate the moles of potassium hydroxide needed, use the formula: moles = molarity * volume (in liters). First, convert 300 mL to liters (0.3 L). Then, moles = 0.250 mol/L * 0.3 L = 0.075 moles of potassium hydroxide needed to prepare the solution.
How would you prepare a 300 ml of a 0.1 M Na CL solution from a solid Na CL solution and water?
You prepare a solution by dissolving a known mass of solute into a specific amount of solvent. In solutions, M is the molarity, or moles of solute per liter of solution. For 300 ml of a 0.1 M Na CL solution from a solid Na CL solution and water you need water and sodium chloride.
How many grams of sodium chloride is in 0.40 moles?
Molarity = moles of solute/Liters of solution ( 300 ml = 0.300 Liters ) For our purposes, Moles of solute = Liters of solution * Molarity Moles NaCl = 0.300 Liters * 0.15 M = 0.05 moles NaCl =============
Please show work How many moles would there be in 300 ml of of a 1.5 M solution?
To find the number of moles, use the formula: moles = Molarity (M) x Volume (L). First, convert 300 ml to liters by dividing by 1000: 300 ml / 1000 = 0.3 L. Then, calculate moles = 1.5 M x 0.3 L = 0.45 moles. Therefore, there would be 0.45 moles in 300 ml of a 1.5 M solution.
How many sugar molecules are present in 300 mL of a 2.0 molar solution?
By the definition of molarity, a 2.0 molar solution contains twice Avodagro's Number of molecules per liter. Therefore, the stated solution contains 0.300X2X6.022Xten to the 23rd power = 3.6X ten to the twenty-third power sugar molecules. (Only two significant digits are justified, because the molarity is specified with only two significant digits.)
What mass of sucrose should be combined with 472 of water to make a solution with an osmotic pressure of 8.95 at 300?
I hate a little trouble with this one myself until I I figured out why in the hell they gave me density. Here is the work: (minus the units because if you can't figure that out or find it in a book may God save your sorry soul.) To start, find the Molarity: 8.95/(.08206*300)= M now realize a couple things... First, being that Molarity is mol (solute)/Liters (SOLUTION). The solution is the water and the sucrose so you can't just plug in the volume of the water and find moles of the solute. Secondly the give you density for the solution which is the same as the solvent (it doesn't say it in the question but its in the actually question given in the homework and probably why no one has answered this yet). The density of the solvent is 1.00 g/mL (density of water. again, God save your soul if you don't know it or couldn't find it) Now since we know that there are there X moles of sucrose in 1 Liter of solution (the Molarity you got) we find how many grams are there in one liter of solution. Let's say M = .345 then you will take that times the molar mass of sucrose and find the grams per Liter. Once you have this number you can then proceed to take that number from 1000 since the density (here where this bit of info finally comes in) is 1 g/mL, ever Liter of solution is 1 kg of solution. Now you do some basic math and set 2 fractions equal to each other, one being X (grams of sucrose) over your grams of water that gave you in the problem and the other being the the amount of grams you found from your Molarity over the amount of water you found you had in 1kg of solution. And there it is just plug in whatever numbers you got and you will get the answer.
What quantity of a 80 percent acid solution must be mixed with a 30 percent solution to produce 300 mL of a 40 percent solution?
Let Q be the quantity of 80% solution required then (300 - Q) is the quantity of 30% solution as together the two solutions must equal 300. Then, [80 x Q] + [30 x (300 - Q)] = [40 x 300] 80Q + 9000 - 30Q = 12000 50Q = 3000 Q = 60...........which means (300 - Q) = 240 60ml of 80% acid solution + 240ml of 30% solution produces 300ml of 40% solution.
How many moles of NaOH needed to prepare 300 mL of a 0.2 m solution of NaOH?
To find the moles of NaOH needed, use the formula: moles = concentration (molarity) x volume (liters). First, convert 300 mL to liters (0.3 L). Then, calculate: moles = 0.2 mol/L x 0.3 L = 0.06 moles. Therefore, 0.06 moles of NaOH are needed to prepare 300 mL of a 0.2 M solution.
