Question

calculate the mass of pottasium sulphate required to prepare 30% (mass percent) solution in 100g of water​

Answer 1

Given :

▪ Mass percent of solution = 30%

▪ Mass of water = 100g

To Find :

▪ Mass of potassium sulphate required to prepare this solution.

Concentration of Solution :

❇ Name : Mass percentage

❇ Symbol : %(w/W)

❇ Definition : Amount of solute in grams present in 100g of solution.

❇ Effect of temperature : No effect

❇ Formula :

$\bigstar\:\underline{\boxed{\bf{\red{\%\dfrac{w}{W}=\dfrac{Mass\:of\:solute}{Total\:mass\:of\:solution}\times 100}}}}$

Calculation :

$\dashrightarrow\sf\:\%\dfrac{w}{W}=\dfrac{W_{K_2SO_4}}{W_{K_2SO_4}+W_{H_2O}}\\ \\ \dashrightarrow\sf\:\dfrac{30}{100}=\dfrac{W_{K_2SO_4}}{W_{K_2SO_4}+W_{H_2O}}\\ \\ \dashrightarrow\sf\:0.3W_{K_2SO_4}+(0.3\times 100)=W_{K_2SO_4}\\ \\ \dashrightarrow\sf\:30=(1-0.3)W_{K_2SO_4}\\ \\ \dashrightarrow\sf\:W_{K_2SO_4}=\dfrac{30}{0.7}\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{W_{K_2SO_4}=42.85\:g}}}}\:\orange{\bigstar}$

Answer 2

Given:

• Percentage of mass in solution = 30%
• Mass of water = 100g

To find:

• Mass percent (w/W %) = ?

Solution:

We know that

w/W % = Mass percent of solute/Total mass of solution × 100

30/100 = W(K2SO4)/W(K2SO4) + W(H2O)

0.3 × 100 + 3[W(K2SO4)] = W(K2SO4)

30 = (1 - 0.3) W(K2SO4)

W(K2SO4) = 30/0.7

W(K2SO4) = 42.85