Determine the molality of a solution containing 0255 g of HNO_3 in 212 g of H_2O

Question

Determine the molality of a solution containing 0255 g of HNO_3 in 212 g of H_2O. Calculate the mole fraction of each component in a solution containing 51.0g of C_6H_6 and 12.0 g CH_3OH. Moles solute/kg solvent is an expression of ? Calculate the number of grams of NaOH in 3.00 L of a 0.390 M solution. How many moles of HCl are present in a solution that has a total mass of 1000g and percent solute 49? A student needs 5.00 mol of NH_3 to perform an experiment. The only available source is a concentrated solution of NH_3 that has the density of 0.90 g/mL and is 29.0% NH_3 by mass. What volume of NH_3 solution should the student use?

Explanation / Answer

4) molality = number of moles / mass (kg)

find number of moles of HNO3 = given mass / molar mass = (0.255 g / 66.01 g/mol) = 0.00386 moles

molality = (0.00386 moles / 0.212 kg ) = 0.0182 m

5) mole fraction = mole of a / total mole

find moles of C6H6 = (51.0 g / 78.11 g/mol) = 0.653 moles

moles of CH3OH = (12.0 g / 32.04 g/mol) = 0.3745 moles

mole fraction of C6H6 = [ 0.653 moles / (0.653 + 0.3745) ] = 0.6355

mole fraction of CH3OH = (1 - 0.6355) = 0.3645

6) molality = moles of solute / mass in kg of solvent

7) molarity = number of moles of solute / volume of solvent (L)

no. of moles of NaOH = 3.0 L x 0.39 M = 1.17 moles

mass of NaOH = 1.17 moles x 39.997 g/mole = 46.79 g

8) 49 % of 1000 g = 490 g

moles of HCl = 490 g / molar mass of HCl = (490 g / 36.46 g/mol) = 13.44 moles

9) density of solution = 0.9 g/ml

let take 100 ml of solution than solute will be = (0.9 g/ml x 100 ml) = 90 g

NH3 is only 29.0 % by mass so

29.0% of 90 g = (90/100) x 29 % = 26.1 g

moles of 26.1 g NH3 = (26.1g / 17.031 g/mol) = 1.5325 moles

1.5325 moles of NH3 from 100 ml of solution

then 5 mole of NH3 = (100 ml / 1.5325 moles) x5 moles = 326.26 mL

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