Question

2. A solution is made by dissolving 9 g of common salt in water so as to get 1 point
150 g of solution. The mass by mass percentage of the solution is:​

Answer 1

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$\bigstar\large\boxed{\mathtt{\green{Question:}}}$

2. A solution is made by dissolving 9 g of common salt in water so as to get 1 point  150 g of solution. The mass by mass percentage of the solution is:​

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$\bigstar\large\boxed{\mathtt{\blue{Given:}}}$

• mass of solute = 9 g
• mass of solution = 150 g

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$\bigstar\large\boxed{\mathtt{\orange{To\: find:}}}$

• mass percentage of solution

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$\bigstar\large\boxed{\mathtt{\pink{Solution:}}}$

As per the formula ,

$\implies\mathtt{\%\dfrac{m}{m} = \dfrac{mass\:of \:solute}{mass\: of \:solution} \times100}$

$\implies\mathtt{\%\dfrac{m}{m} = \dfrac{9}{150} \times100}$

$\implies\mathtt{\% \dfrac{m}{m} = 0.06 \times 100}$

$\implies\mathtt{\red{\%\dfrac{m}{m} = 6 \%}}$

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$\bigstar\large\boxed{\mathtt{\pink{Additional \: formulas :}}}$

$\implies\mathtt{\%\dfrac{m}{v} = \dfrac{mass\:of \:solute}{volume\: of \:solution} \times100}$

$\implies\mathtt{\%\dfrac{v}{v} = \dfrac{volume\:of \:solute}{volume\: of \:solution} \times100}$

$\implies\mathtt{molarity=\dfrac{moles \:of \:solute}{vol\: of \:solution(L)} }$

$\implies\mathtt{molality=\dfrac{moles \:of \:solute}{mass\: of \:solvent(kg)} }$