Question

A 9 litre solution of a dilute hydrochloric acid has a strength of 50% to reduce the strength to 30% the amount of a water needed is
A) 3 litre B) 4litre C) 5litre D) 6litre explain answer​

Answer 1

As Given :-

Initial concentration = ${C}_{1}$ = 50% = 50/100 = 1/2

Final Concentration = ${C}_{2}$ = 30% = 30/100 = 3/10

Initial volume = ${V}_{1}$ = 9 L

As per relations we know that,

${C}_{1}{V}_{1} =\:{C}_{2}{V}_{2}$

1/2 × 9 = 3/10 × ${V}_{2}$

${V}_{2}$ = 15 L

The final volume includes initial volume also.

Volume of Water needed = (15 - 9)L = 6 L

Answer 2

Given:-

• A 9 litre solution of a dilute hydrochloric acid has a strength of 50% to reduce the strength to 30%.

To Find:-

• Amount of a water needed

Formula Used:-

• ${\boxed{\bf{C_I V_I = C_F V_F}}}$

Here,

• $\bf C_I=$ Initial Concentration

• $\bf C_F=$ Final Concentration

• $\bf V_I=$ Initial Volume

• $\bf V_F=$ Final Volume

Solution:-

Firstly,

• $\bf C_I=50 \% = \dfrac{50}{100}= \dfrac{1}{2}$

• $\bf C_F=30 \% = \dfrac{30}{100}= \dfrac{3}{10}$

• And, here $\bf V_I=9\:L$

Using, $\bf C_I V_I = C_F V_F$

$\sf :\implies\: \dfrac{1}{2}\times 9 = \dfrac{3}{10} \times V_F$

$\sf :\implies\: V_F= \dfrac{9\times10}{2\times3}$

$\sf :\implies\: V_F= \dfrac{10\times3}{2}$

$\sf :\implies\: V_F= 15\:L$

$\bf Amount\: of\: a\: water \:needed \:=V_F - V_I$

$\bf Amount\: of\: a\: water \:needed \:=15 - 9$

$\bf Amount\: of\: a\: water \:needed \:=6\:L$

Hence, the amount of a water needed is 6 litres.

━━━━━━━━━━━━━━━━━━