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Standard XII

Chemistry

Molarity

10.0 grams of...

Question

$10.0 g r a m s$ of caustic soda when dissolved in $250 {c m}^{3}$ of water, the resultant molarity of solution is:

0.25 M

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0.5 M

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1.0 M

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0.1 M

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Solution

The correct option is D $1.0 M$
Caustic soda implies NaOH, molar mass of which is 40 grams
Thus, when 40g of NaOH is dissolved in 1 litre(1000 $c m^{3}$ ) of water, resulting solution is 1M
i:e $40 g = 1000 \times 1 M$
So for 10g,
= $\frac{1000 \times 10 \times 1}{40 \times 250} = 1 M$

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(a)

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g of caustic soda is dissolved in

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mL of the solution.
(b)

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mL of solution.
(c)

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g of pure HCl gas is dissolved in

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is dissolved in

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10.0 grams of caustic soda when dissolved in 250 cm3 of water, the resultant molarity of solution is:

The correct option is D 1.0 MCaustic soda implies NaOH, molar mass of which is 40 gramsThus, when 40g of NaOH is dissolved in 1 litre(1000 cm3) of water, resulting solution is 1Mi:e 40g=1000×1MSo for 10g,=1000×10×140×250=1M

$10.0 g r a m s$ of caustic soda when dissolved in $250 {c m}^{3}$ of water, the resultant molarity of solution is:

The correct option is D $1.0 M$
Caustic soda implies NaOH, molar mass of which is 40 grams
Thus, when 40g of NaOH is dissolved in 1 litre(1000 $c m^{3}$ ) of water, resulting solution is 1M
i:e $40 g = 1000 \times 1 M$
So for 10g,
= $\frac{1000 \times 10 \times 1}{40 \times 250} = 1 M$