When 100 mL of 80 % (w/v) NaOH is mixed with certain amount of 40 %( w / w ) NaOH solution of density 1.2 g / mL, the percentage concentration of final solution becomes 60 % (w/v). Find the molarity of the final solution.

Question

When 100 mL of 80 % (w/v) NaOH is mixed with certain amount of 40 %( w / w ) NaOH solution of density 1.2 g / mL, the percentage concentration of final solution becomes 60 % (w/v). Find the molarity of the final solution. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

0.6=(80+V × 1.2 × 0.4/100+V) 60+0.6 V=80+0.48 V 0.12 V=20 V=(1000/6) mL Total mass of NaOH mixed =80+V × 1.2 × 0.4 =80+(1000/6) × 1.2 × 0.4=160 g . Molarity =15 M

Q. When $100 m L$ of $80%$ (w/v) $N a O H$ is mixed with certain amount of $40% (w / w) N a O H$ solution of density $1.2 g / m L$ , the percentage concentration of final solution becomes $60%$ (w/v). Find the molarity of the final solution.

$0.6 = \frac{80 + V \times 1.2 \times 0.4}{100 + V}$
$60 + 0.6 V = 80 + 0.48 V$
$0.12 V = 20$
$V = \frac{1000}{6} m L$
Total mass of $N a O H$ mixed $= 80 + V \times 1.2 \times 0.4$
$= 80 + \frac{1000}{6} \times 1.2 \times 0.4 = 160 g .$
Molarity $= 15 M$

Q. When 100mL of 80% (w/v) NaOH is mixed with certain amount of 40%(w/w)NaOH solution of density 1.2g/mL, the percentage concentration of final solution becomes 60% (w/v). Find the molarity of the final solution.

0.6=100+V80+V×1.2×0.4​ 60+0.6V=80+0.48V 0.12V=20 V=61000​mL Total mass of NaOH mixed =80+V×1.2×0.4 =80+61000​×1.2×0.4=160g. Molarity =15M

Q. When $100 m L$ of $80%$ (w/v) $N a O H$ is mixed with certain amount of $40% (w / w) N a O H$ solution of density $1.2 g / m L$ , the percentage concentration of final solution becomes $60%$ (w/v). Find the molarity of the final solution.

$0.6 = \frac{80 + V \times 1.2 \times 0.4}{100 + V}$
$60 + 0.6 V = 80 + 0.48 V$
$0.12 V = 20$
$V = \frac{1000}{6} m L$
Total mass of $N a O H$ mixed $= 80 + V \times 1.2 \times 0.4$
$= 80 + \frac{1000}{6} \times 1.2 \times 0.4 = 160 g .$
Molarity $= 15 M$