Question

1. Calculate the number of moles of sulfuric acid that is contained in 250 mL of 8.500 M sulfuric acid solution


2.7.300 moles of sodium nitrite are needed for a reaction. The solution is 5.450 M. How many mL are needed?


3.What mass (in g) of NH3 must be dissolved in 875 g of methanol to make a 0.430 molal solution?


4.Calculate the molality of a solution that is prepared by mixing 259.5 mL of CH3OH

(d = 0.792 g/mL) and 1387 mL of CH3CH2CH2OH (d = 0.811 g/mL)


6.A solution is prepared by dissolving 318.6 g sucrose (C12H22O11) in 4905 g of water. Determine the molarity of the solution

Answer 1

1.n=C*V=0.25*8.5=2.125 mol

2.V=n/C=7.3/5.45=1.34 L

3.m(NH3 )=C_m*m(CH3OH)*Mr(NH3)=0.43*0.875*17=6.4 g

4.C_m=V(CH3OH)*d(CH3OH)*1000/(Mr(CH3OH)*V(CH3CH2CH2OH)*d(CH3CH2CH2OH))=259.5*0.792*1000/(32*1387*0.811)=5.71 mol/kg

6.C_M=m(C12H22O11)/Mr(C12H22O11)*V(H2O)=318.6/(342*4.905)=0.19 mol/L