Question

34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality and mole fraction of sugar in the syrup. Given C = 12 and O = 16

Answer 1

Answer:

1). Mole fraction of sugar in syrup = 1/101

2). Molality of sugar in syrup = 0.5 mol/ kg

>> NOTE : MOLARITY CAN'T BE CALCULATED WITH GIVEN MASS, IT IS ONLY CALCULATED WITH VOLUME, BUT MOLALITY CAN BE CALCULATED.

Explanation:

1). Mole fraction :

>>no. of moles of sugar(solute)/total moles of solution(solution)

(1). sugar = C12H22011

Weight of sugar = 34.2g (given)

Molar mass of sugar = 342 g/mol

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>> mole of sugar = 34.2/342 = 1/10 mole

(II). Water = H2O

Weight of water = 214.2-34.2=180 g Molar mass of water = 18 g/mol

>> mole of water = 180/18 = 10 mole

Now,

>> Mole fraction = (1/10)/(101/10)

= 1/10 × 10/101

[Answer] = 1/101 [ Mole fraction

is a unit less quantity]

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2). Molality

>> If mass is given in gram then,

Molality (weight of solute × 1000)/ = (Molar mass of solute × weight of solvent)

Hence,

weight of solute (Sugar) = 34.2 g (Given)

weight of solvent (water) = 180 g (Given)

Molar mass of solute = = 342 g

» (34.2 x 1000)/(342 × 180)

= 34200/61560

= 0.5 mole/Kg [Answer]

Answer 2

GIVEN:-

• Mass of solute
• Molar mass of sugar

TO FIND:-

Molality and mole

EXPLANATION:-

(1)We know:-

$\sf\;Molality=\frac{Number\;of\;mole\;of\;solution}{Weight\;of\;solvent\;in\;kg}$

$\sf\;Number\;of\;moled\;of\;sugar=\frac{34.2}{342}\\\\\rightarrow 0.1$

Now let's calculate mass of water,

Mass of water=214.2-34.2

Mass of water=180 gm

Mass of water >180/1000 kg

$\sf\;Molality=\frac{Number\;of\;moles\;of\;sugar}{Moles\;of\;water\;in\;kg}\\\\\rightarrow Molality=\frac{0.1}{180}\times 1000\\\\==>0.555\;M$

So the molality is 0.555 M

So total number of moles=10+0.1

==>10.1

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(2)

$\sf\;Mole\;fraction\;of\;sugar=\frac{Number\;of\;moles\;of\;sugar}{Total\;number\;of\;moles}\\\\\rightarrow \frac{0.1}{10.1}\\\\==>0.0099$

So the total mole fraction of syrup is 0.0099

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