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What volume of a 0.150 M saltwater solution conatins 0.500 g of Nacl?
To find the volume of a 0.150 M saltwater solution that contains 0.500 g of NaCl, first convert grams of NaCl to moles using its molar mass (approximately 58.44 g/mol). 0.500 g NaCl × (1 mol NaCl / 58.44 g) ≈ 0.00857 mol NaCl. Next, use the molarity formula (M = moles/volume in liters) to find the volume: Volume = moles / Molarity = 0.00857 mol / 0.150 M ≈ 0.0571 L, or 57.1 mL.
What is the molarity of 750 ml of NaCl solution that contains 21.04g of NaCl?
To calculate the molarity (M) of the NaCl solution, first find the number of moles of NaCl. The molar mass of NaCl is approximately 58.44 g/mol, so 21.04 g of NaCl is about 0.360 moles (21.04 g ÷ 58.44 g/mol). Molarity is defined as moles of solute per liter of solution; therefore, for 750 ml (0.750 L), the molarity is 0.360 moles ÷ 0.750 L = 0.480 M. Thus, the molarity of the NaCl solution is 0.480 M.
What is the concentration in (M) when 34.6 grams of NACL is added 1500 ml of water?
To find the concentration in molarity (M), first calculate the number of moles of NaCl. The molar mass of NaCl is approximately 58.44 g/mol, so 34.6 grams of NaCl is about 0.592 moles (34.6 g / 58.44 g/mol). Then, divide the number of moles by the volume of the solution in liters: 0.592 moles / 1.5 L = 0.395 M. Therefore, the concentration is approximately 0.395 M.
How many milliliters of 2 M NaCl solutions are required to make 1 liternof 0.4 M NaCl solution?
To prepare 1 liter of a 0.4 M NaCl solution from a 2 M NaCl solution, you can use the dilution formula, (C_1V_1 = C_2V_2), where (C_1) is the concentration of the stock solution, (V_1) is the volume of the stock solution needed, (C_2) is the concentration of the diluted solution, and (V_2) is the final volume of the diluted solution. Plugging in the values: (2 , \text{M} \times V_1 = 0.4 , \text{M} \times 1000 , \text{mL}). Solving for (V_1) gives (V_1 = \frac{0.4 \times 1000}{2} = 200 , \text{mL}). Thus, 200 mL of the 2 M NaCl solution is required.
What is is the molality of 3.00 of NaCL and 1.50kg?
m = moles/kg = 3.00/1.50 = 2.00 molal
What is 5 M of NaCl equivalent to?
5 M of NaCl is equivalent to 292,2 grams.
What is the total dilution factor from 1.00 M NaCl to 1.00 M NaCl solution?
The concentration is the same !
preparation of 0.1 m of NaCL ML SOLUTION AND ITS LAP REPORT?
A solution of NaCl 1 M.
Calculate the isotonic coefficient for NaCl if a 07M of NaCl solution equaled the hemolysis of14M glucose solution?
i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2
How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?
To find the volume of 0.075 M NaCl solution that can be made, you can use the formula C1V1 = C2V2. Plugging in the values, we get (9.0 M)(450 mL) = (0.075 M)(V2). Solving for V2 gives V2 = 1.1 L. Therefore, 1.1 L of 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl.
5 percent NaCl equals x M NaCl?
5 % NaCl is equal to 5 g NaCl in 100 g of a material.
Which will have the lowest boiling point pure water 0.5 M NaCl aq 1.0 M NaCl aq or 2.0 M NaCl aq?
Pure water will have the lowest boiling point because it does not contain any solute particles to elevate the boiling point. As the concentration of NaCl increases, the boiling point also increases due to an increase in the number of solute particles that disrupt the formation of water vapor. Therefore, 0.5 M NaCl will have a higher boiling point than pure water, followed by 1.0 M NaCl, and finally 2.0 M NaCl will have the highest boiling point.
How many moles per mL of the solute are contained in the 0.20 M NaCl?
0.0002 mol NaCl/mLmolarity = (moles solute)/(L of solution)0.20 M NaCl = 0.20 mol NaCl/L1 L = 1000 mL0.20 mole NaCl/1000 mL = 0.00020 mol NaCl/mL (rounded to two significant figures)
You want 1 liter of 0.1 M NaCl and you have 4 M stock solution How much of the 4 M solution and how much dH2O will you measure out for this dilution?
To make 1 liter of 0.1 M NaCl solution, you will need 25 ml of the 4 M NaCl stock solution and 975 ml of water. This will give you the desired concentration of 0.1 M NaCl in a total volume of 1 liter.
What is the mole concentration of 58.44g of NaCl in 1 L of water?
Molarity = moles of solute/Liters of solutionGet moles NaCl.58.44 grams NaCl (1 mole NaCl/58.44 grams)= 1 mole NaCl------------------Molarity = 1 mole NaCl/1 liter= 1 M NaCl========
After titrating 0.1 M HCl with 0.1 M NaOH what material remain in the flask?
After the titration of 0.1 M HCl with 0.1 M NaOH, the material remaining in the flask would be sodium chloride (NaCl) along with water. The reaction between HCl and NaOH forms water and NaCl as the products.
How much of NaCl is in 1.84 L of 0.200 M NaCl Answer in units of mol.?
To find the amount of NaCl in 1.84 L of 0.200 M NaCl, you use the formula: amount = concentration x volume. Thus, amount = 0.200 mol/L x 1.84 L = 0.368 mol of NaCl.
