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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150M+ Students 30,000+ Colleges 500+ Exams 1500+ E-books 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? No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections 1 |