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 = mol v

The moles of sodium chloride is 5.85 58.4 0.1 mol

The volume of water is 0.5 l

So

M = mol v = 0.1 0.5 = 0.2 M

Cheers!

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

Text Solution

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