Mass Percentage Problem 3

Initially,
Mass of $NaOH = \displaystyle\frac{10}{100}\times 50 = 5g$

Let's assume we added $mg$ of $30$ % $NaOH$ solution, then,

Mass of $NaOH = 5+\displaystyle\frac{30}{100}\times m = (5+0.3m)g$
Mass of solution finally $= (50+m)g$

Thus, since we want to make a $15$ % solution,

$\displaystyle\frac{5+0.3m}{50+m}\times 100 = 15$

Solving, we get,

$m=\displaystyle\frac{50}{3} \approx \boxed{16.67g}$

With 50 g as a basic as 100% or 1,

g ( $H_2O$ ), g ( $NaOH$ ), %:

1
2

45  5   10
35  15  30

With fraction x of 0 < x < 1 of 50 g,

$\frac{5 + x 15}{50 + x 50} = 15$ % (where $\frac{5 + 15}{50 + 50} = 20$ % is logical with x = 1.)

$\frac{5(1 + 3 x)}{50 (1 + x)} = \frac{3}{20}$ such that 10% $(1 + 3 x)$ = 15% $(1 + x).$

$\frac{(1 + 3 x)}{(1 + x)} = \frac32$ else think of (1) 10% + (x) 30% = (1 + x) 15% as simplified way.

2 + 6 x = 3 + 3 x $\implies$ x = $\frac13$

$\frac{50 g}{3}$ of 30% NaOH is required. (Not taking 35 g of $H_2O$ of the 30% into calculation.)

Logical verification: $\frac{5 + \frac{15}{3}}{50 + \frac{50}{3}} = \frac{30}{200} = 15$ % {Correct!}

Answer: $\boxed{16.7 g}$