Q.9.57 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine the final volume, in millilitres, of each of the following:

a. a 1.5 M HCL solution prepared from 20.0mL of a 6.0M HCl solution

b. a 2.0% (m/v) LiCl solution prepared from 50.0mL of a 10.0% (m/v) LiCl solution

c. a 0.500M${\mathrm{H}}_3{\mathrm{PO}}_4$solution prepared from 50.0mL of a 6.00M ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution

d. a 5.0% (m/v) glucose solution prepared from 75ml of a 12 % (m/v) glucose solution

Step-by-Step Solution

Answer

a. The final volume in millilitres is $80$ml

b. The final volume in millilitres is $250$ml

c. The final volume in millilitres is $600$ml

d. The final volume in millilitres is$180$ml

Part (a) Step 1: Given Information

a $1.5$M HCL solution prepared from $20$mL of a $6$M HCl solution 

Part (a) Step 2: Explanation

Given,

$C_1=6\mathrm{M}, {\mathrm{V}}_1=20\mathrm{mL}, {\mathrm{C}}_2=1.5\mathrm{M}$

Using the concentration formula,

$C_1{V}_1=C_2{V}_2$

The final volume in millilitres is 

$V_2=\frac{C_1{V}_1}{C_2}=20mL\times \frac{6\mathrm{M}}{1.5M}=80mL$

Part (b) Step 1: Given Information

a $2$% (m/v) LiCl solution prepared from $50$mL of a $10$% (m/v) LiCl solution 

Part (b) Step 2: Explanation

Given,

${\mathrm{C}}_1=10\mathrm{\%}(\mathrm{M}/\mathrm{V}), {\mathrm{V}}_1=50\mathrm{mL}, {\mathrm{C}}_2=2\mathrm{\%}(\mathrm{M}/\mathrm{V})$

Using the concentration formula,

$C_1{V}_1=C_2{V}_2$

The final volume in millilitres is 

$V_1=\frac{C_2{V}_2}{C_1}=50mL\times \frac{10\mathrm{\%}}{2\mathrm{\%}}=250mL$

Part (c) Step 1: Given Information

a $0.500$M ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution prepared from $50$mL of a $6.00$M ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution 

Part (c) Step 2: Explanation

Given,

$C_1=6\mathrm{M}, V_1=50\mathrm{mL}, C_2=0.5\mathrm{M}$

Using the concentration formula,

$C_1{V}_1=C_2{V}_2$

The final volume in millilitres is 

$V_2=\frac{C_1{V}_1}{C_2}=50mL\times \frac{6M}{0.5M}=600mL$

Part (d) Step 1: Given Information

a $5$% (m/v) glucose solution prepared from $75$ml of a $12$% (m/v) glucose solution 

Part (d) Step 2: Explanation

Given,

$C_1=12\mathrm{\%}(M/V), V_1=75\mathrm{mL}, C_2=5\mathrm{\%}(M/V)$

Using the concentration formula,

$C_1{V}_1=C_2{V}_2$

The final volume in millilitres is  

$V_1=\frac{C_2{V}_2}{C_1}=75\mathrm{mL}\times \frac{12\mathrm{\%}}{5\mathrm{\%}}=180\mathrm{mL}$