$4 L$ of $0.02 M$ aqueous solution of $N a C l$ was diluted by adding $1 L$ of water. The molarity of the resultant solution is:

0.004

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0.008

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0.012

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0.016

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Solution

The correct option is B $0.016$
Since, $M_{1} V_{1} = M_{2} V_{2}$

Let $M_{1}$ and $V_{1}$ be the molarity and volume of the given solution.

And $M_{2}$ and $V_{2}$ be the molarity and volume of the resulting solution.

$M_{1} = 0.02 M$

$V_{1} = 4 L$

$V_{2} = 5 L$

Therefore, $0.02 \times 4 = M_{2} \times 5$

$M_{2} = 0.016 M$

Hence, option D is correct.

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Question 24
4 L of 0.02 M aqueous solution of NaCl was diluted by adding 1 L of water. The molality of the resultant solution is ........ .
(a) 0.004 (b) 0.008 (c) 0.012 (d) 0.016

4L of 0.02M aqueous solution of NaCl was diluted by adding 1L of water. The molality of the sol. Is

Q. 100 mL of a 0.1 M NaCl solution is added to 300 mL of a 0.2 M

M g C l_{2}

solution. 100 mL of this solution was further diluted with 400 mL of water. The molarity of

C l^{-}

ions in the resultant solution is:

100 mL

of a

0.1 M N a C l

solution is added to

300 mL

of a

0.2 M

M g C l_{2}

solution.

100 mL

of this solution was further diluted with

400 mL

of water. The molarity of

C l^{-}

ions in the resultant solution is:

4 L

of water is added to

6 L

of a

2 M

H C l

solution. What is the molarity of the resultant solution?

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4L of 0.02M aqueous solution of NaCl was diluted by adding 1L of water. The molarity of the resultant solution is:

The correct option is B 0.016Since, M1V1=M2V2Let M1 and V1 be the molarity and volume of the given solution.And M2 and V2 be the molarity and volume of the resulting solution.M1=0.02MV1=4LV2=5L Therefore, 0.02×4=M2×5M2=0.016MHence, option D is correct.

$4 L$ of $0.02 M$ aqueous solution of $N a C l$ was diluted by adding $1 L$ of water. The molarity of the resultant solution is: