Solution, Solute and Solvent Problem 5

The ratio for $NaCl$ to dissolve in water is $36g:100g$ water

So we can use the same ratio to solve this problem

We can both sides by $36$ and multiply both sides by $40$

$\frac{36}{36}\times 40:\frac{100}{36}\times 40$

And we can get $40g:111.1111g$

Rounded to the nearest ones is $\boxed{111g}$

Solubility of $NaCl$ in $xg$ of water $= \frac{36}{100} x g$

Now, given is that $40g$ of $NaCl$ is used.

Hence,
$\frac{36}{100} x = 40g$
$\Rightarrow x = 111.11g \approx \boxed{111 g}$

I did by unitary system, :)

36gm of salt is dissolved in 100 gm of water

1 gm of salt is dissolved in 100/36 gm of water

Therefore, 40 gm of salt will be dissolved in 100X40/36 = 111.11gm of water

Let NaCl be X and Water be Y. Then 36X= 100Y (saturated solution). Then, 36X x 40/36 = 100Y x 40/36 ( new saturation point). Hence, 40X= 111Y Therefore, it is 111g.

36/100 = 40/x. Thus, x = 40/36 100 = 10 11.111111... = 111.1111111 = 111(approx.)