Solutions containing 23g HCOOH is / are 

$\text{ 1) }46g\text{ of }70\mathrm{\%}\frac{w}{v}d=1.40g/mL$

$46\times \frac{70}{140}=23g$

$\text{ 2) }M=\frac{w}{GMw}\times \frac{1000}{V}{d}_{soln}=gm/mL$

$10=\frac{w}{46}\times \frac{1000}{50}$

$\text{ 3) }25g\to 100g$

$\frac{25\times 50}{100}\leftarrow 50g$

$=12.5g$

$\text{ 4) }5=\frac{w}{46}\times \frac{1000}{46}$

$w=10.58g$