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First, find the molecular weights of the individual elements. Next, multiply each of the weights of the individual elements in the NaOH. Next, sum the molecular weights. What does the resulting number give you? That is, what does it express? Does it provide one of the entities needed to determine molarity? Next, determine how many 250-ml volumes are in a liter. You now have the two entities necessary to calculate the answer to your question. Hop to it.
40 grams, this is the 1M NaOH standard laboratory solution.
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
You need 2,4 g NaOH (0,06 moles).
First, find the molecular weights of the individual elements. Next, multiply each of the weights of the individual elements in the NaOH. Next, sum the molecular weights. What does the resulting number give you? That is, what does it express? Does it provide one of the entities needed to determine molarity? Next, determine how many 250-ml volumes are in a liter. You now have the two entities necessary to calculate the answer to your question. Hop to it.
40 grams, this is the 1M NaOH standard laboratory solution.
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
You need 2,4 g NaOH (0,06 moles).
Molarity = moles of solute/Liters of solution. ( 350 ml = 0.350 Liters ) 5.7 M NaOH = moles NaOH/0.350 Liters = 1.995 moles NaOH (39.998 grams/1 mole NaOH) = 78 grams NaOH needed ------------------------------------
Dissolve 0.4 g of NaOH in 100 ml of water. Try it out. Actually it is not suitable to prepare NaOH solutions in standard flasks.It should be made in beakers & must be standardised..This is done to find the correct normality...
112,64 g of NaOH and 138,1 g of H2SO4
molarity equals moles of solute /volume of solution in litres . moles of NaOH equals 5g/40g = 0.125 and volume of solution will be volume of water + volume of NaOH = 0.5 litre+0.002 l which is nearly 0.5 litre . (volume of NaOH is calculated by its density) so molarity = 0.125mol/0.5litre = 0.25 M
80g of NaOH dissolved in 250ml. of water find the molarity of this solution ?
First, you must find the amount of moles of NaOH, using the concentration and volume given. By lowercase m, I'm assuming you mean molality, or molals of solution, which is the equation:molality (m) = (moles of solute) / (total volume of solution (in liters))To solve for moles of NaOH, your solute, rearrange the equation by multiplying volume on both sides to get:moles solute = (molality)(total volume of solution)Next, just plug in the information you know, which is 500 mL for the total volume and 125 m for the molality.***Volume for concentration problems must be converted to liters, so remember that 1 L = 1000 mLmoles NaOH = (125 m)(0.500 L) = 62.5 molesFinally, convert this to grams by finding the molar mass of NaOH using the periodic table:22.99 + 16.00 + 1.008 = 39.998 g/mol62.5 moles (39.998 g) / (1 mol) =249.875 grams NaOH
8 grams NaOH (1 mole NaOH/39.998 grams) = 0.2 moles NaOH