Question

A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.

Answer 1

Data:
$m_{1} = 2\:g$
$m_{2} = 18\:g$
$\tau (title) = ?$
$\tau\% (the\:mass\:per\:cent\:of\:the\:solute) = ?$

Solving:
$\tau = \frac{ m_{1} }{m} \to\:\tau = \frac{ m_{1} }{m_{1}+ m_{2} }$
$\tau = \frac{2}{2+18}$
$\tau = \frac{2}{20}$
$\boxed{\tau = 0.1}$

the mass per cent of the solute:
$\tau\%= \tau* 100\%$
$\tau\% = 0.1*100\%$
$\boxed{\boxed{\tau\% = 10\%}}\end{array}}\qquad\quad\checkmark$

Answer 2

HEYA MATE HERE U GO.

Mass percent of A = Mass of A / Mass of solution × 100.

= 2g / 2g of A + 18 g of water × 100 = 2g/20g× 100 = 10%....

HOPE THIS WILL HELP UH.