Question

Calculate the concentration in terms of mass percent of a solution obtained by dissolving 25 g of sugar in 150g of water?

Answer 1

AnswEr :

Mass percentage of the sugar is 14.28

Here,

• Sugar is the solute

• Water is the solvent

Given,

• Mass of Solute = 25 g

• Mass of Solvent = 150 g

We have to find the mass percentage of the sugar in the resulting solution

Mass Percentage is given as :

$\star \: \boxed{ \boxed{\bigg( \sf\dfrac{m}{M} \bigg) \%\: = \dfrac{M_{solute}}{ \:M_{solution}} \times 100}}$

Now,

Mass of the Solution = Mass of Solute + Mass of Solvent

» M = (25 + 150) g

» M = 175 g

Mass of the solution is 175 g

Thus,

$\bigg( \sf\dfrac{m}{M} \bigg) \%\: = \dfrac{25}{175} \times 100 \\ \\ \longrightarrow \: \sf \: \bigg(\dfrac{m}{M} \bigg)\% = \dfrac{1}{7} \times 100 \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \sf \: \bigg(\dfrac{m}{M} \bigg)\% = 14.28}}$

Answer 2

Answer:

• The Mass percent (M %) is 14.285 %

Given:

1. Mass of Sugar = 25 grams.
2. Mass of water = 150 grams.

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\bullet\;{\underline{\boxed{\sf M\;\%=\Bigg\lgroup\dfrac{M_{1}}{M}\Bigg\rgroup\times 100}}}$

Here,

• M% Denotes Mass percent.
• M₁ Denotes Mass of sugar (Solute)
• M Denotes Total Mass.

Substituting the values,

$\displaystyle \Rightarrow \sf M\;\%=\Bigg\lgroup\dfrac{M_{1}}{M_1+M_2}\Bigg\rgroup\times 100\ \ \because(M=M_1+M_2)\\\\\\\Rightarrow \sf M\;\%=\Bigg\lgroup\dfrac{25}{25+150}\Bigg\rgroup\times 100\\\\\\\Rightarrow\sf M\;\%=\Bigg\lgroup\dfrac{25}{175}\Bigg\rgroup\times 100\\\\\\\Rightarrow\sf M\;\%=\Bigg\lgroup 0.14285\Bigg\rgroup\times 100\\\\\\\Rightarrow\sf M\;\%=0.14285\times100\\\\\\\Rightarrow \large{\underline{\boxed{\red{\sf M\;\%=14.285\;\%}}}}$

∴ The Mass percent (M %) is 14.285 %

Note:

• Mass of solution = Mass of solute (Sugar) + Mass of solvent (Water)

$\rule{300}{1.5}$