Question

Solve and show step by step solution.

1. The density of the solution is 0.982 g/mL and the density of water is 1.00

g/ml. If the solution is composed of 30.6 g NH3 in 81.3 g of H2O. Compute for;

a. mole fraction of NH3

b. molarity of NH3

c. molality of NH3

2. Given the reaction, H3PO4 + 3 NaOH Na3PO4 + 3 H2O, what is the

mass of 0.20 M H3PO4 needed to react with 100 ml. of 0.10 M NaOH?

Answer 1

$M(NH_3)=14+3x1=17 g/mol$

$n(NH_3)=30.6/17=1.8 mol$

$M(H_2O)=2x1+16=18 g/mol$

$n(H_2O)=81.3/18=4.52$

a. $w(NH_3)=1.8/(4.52+1.8)=0.28$

b.$V(H_2O)=81.3x1=81.3 ml=0.081 l$

$V(NH_3)=30.6/0.982=31.2 ml=0.0312 l$

$c(NH_3)=1.8/(0.081+0.0312)=16.1 mol/l$

c. $c(NH_3)=1.8/(0.0306+0.0813)=16.1 mol/kg$

2. $n(NaOH)=0.1x0.1=0.01 mol$

$n(H_3PO_4)=0.01/3=0.0033 mol$

$V(H_3PO_4)=0.0033x0.2=0.00067 l$

If dencity of H₃PO₄ is 1 g/ml, then

0.00067l=0.67 ml

0.67x1=0.67g