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Question: > What will be the molarity of a solution, which contains \(5.85g\)of Nacl\((s)\) per 500 mL? > >...

What will be the molarity of a solution, which contains 5.85g5.85gof Nacl(s)(s) per 500 mL?

A

4molL−14molL^{- 1}

B

20molL−120molL^{- 1}

C

0.2molL−10.2molL^{- 1}

D

2molL−12molL^{- 1}

Answer

0.2molL−10.2molL^{- 1}

Explanation

Solution

(3) : Molarity (M)

= No.ofmolesofsoluteVollumeofsolution(inL)\frac{No.ofmolesofsolute}{Vollumeofsolution(inL)}

Or Molarity

=WB×1000MB×Volumeofsolution(inml)\frac{W_{B} \times 1000}{M_{B} \times Volumeofsolution(inml)}

=5.85×10058.5×500= \frac{5.85 \times 100}{58.5 \times 500} (∵MolarmassofNaCl=58.5gmol−1(\because MolarmassofNaCl = 58.5gmol^{- 1}

=0.2MolL−10.2MolL^{- 1}