What quantity (in mL) of a 45 % acid solution of a monoprotic strong acid must be mixed with a 20 % solution of the same acid to produce 800 mL of a 29.875 % acid solution ?

Question

What quantity (in mL) of a 45 % acid solution of a monoprotic strong acid must be mixed with a 20 % solution of the same acid to produce 800 mL of a 29.875 % acid solution ? (A) 320 (B) 325 (C) 316 (D) 330

Tardigrade Admin · Accepted Answer

(V×45/100)+((800-V)20/100)=(800×29.875/100) (9V/20)+160-(V/5)=239 (5V/20)=79 ⇒ V=316

Q. What quantity (in mL) of a $45%$ acid solution of a monoprotic strong acid must be mixed with a $20%$ solution of the same acid to produce $800 m L$ of a $29.875%$ acid solution ?

320

325

316

330

$\frac{V \times 45}{100} + \frac{( 800 - V ) 20}{100} = \frac{800 \times 29.875}{100}$

$\frac{9 V}{20} + 160 - \frac{V}{5} = 239$

$\frac{5 V}{20} = 79$

$\Rightarrow V = 316$

Q. What quantity (in mL) of a 45% acid solution of a monoprotic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution ?

320

325

316

330

100V×45​+100(800−V)20​=100800×29.875​ 209V​+160−5V​=239 205V​=79 ⇒V=316

Q. What quantity (in mL) of a $45%$ acid solution of a monoprotic strong acid must be mixed with a $20%$ solution of the same acid to produce $800 m L$ of a $29.875%$ acid solution ?

$\frac{V \times 45}{100} + \frac{( 800 - V ) 20}{100} = \frac{800 \times 29.875}{100}$

$\frac{9 V}{20} + 160 - \frac{V}{5} = 239$

$\frac{5 V}{20} = 79$

$\Rightarrow V = 316$