What is the molarity of the solution obtained by dissolving 3.42 gram of sucrose in hundred Centimetre cube of the solution?

Hi numan,

As Molarity= moles of solute/ volume of solution(liters)

M= 5.85*1000/((23+35.5)*500)=0.2M

If you have any further query, feel free to ask.

Hello student,

The formula for molarity $M$ is

$M$ $=$ $\frac{mol}{v}$

The moles of sodium chloride is $\frac{5.85}{58.4} \approx 0.1$ $mol$

The volume of water is $0.5$ $l$

$M$ $=$ $\frac{mol}{v} = \frac{0.1}{0.5} = 0.2$ $M$

$Cheers!$

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Text Solution

Solution : `n_("sucrose")=(mass)_("sucrose")/("molar mass")` `=(3.42 g)/(342 g mol^(-1))=0.01 mol` Since the density of water is `1.00 g mL^(-1)`, we can find the mass of water as follows: `d_(H_(2)O)=m_(H_(2)O)/V_(H_(2)O)` or `m_(H_(2)O)=(d_(H_(2)O))(V_(H_(2)O))` `=(1.00 g mL^(-1))(50.0 mL)` `=50.0 g` Thus, the molality of the solution is `Molality=n_("sucrose")/g_(H_(2)O)xx(1000 g)/(kg)` `=(0.01 mol)/(50.0 g)xx(1000 g)/(kg)` `=0.2 m`

Solution of 'A' containing 3.42 g of it in 100 cm3 of water is found to be isotonic with a 6.84% `("w"/"V")` solution of sucrose, what is the molar mass of 'A'? (At. wt: C = 12, H = 1, O = 16)

171

Explanation:

Molecular mass of Sucrose = C12H22O11 = 342 g mol-1

Molarity of Sucrose Solution = `("w" xx 1000)/("M" xx "V"_("cm"^3)) = (6.84 xx 1000)/(342 xx 100)` = 0.2 M

∴ Molarity of 'A' = molarity of sucrose solution= 0.2 M

Molarity of 'A' solution = `("w" xx 1000)/("M" xx 100)`

0.2 = `(3.42 xx 1000)/("M" xx 100)`

M = `34.2/0.2` = 171 g mol-1

Concept: Osmotic Pressure

Is there an error in this question or solution?

Molarity of a solution that has 3.42 g sucrose in 500 mL solution is.A. 0.2B. 0.02C. 2D. 0.342

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