Question

The molality of a solution containing 60 g of glucose ( C6 H12 O6 ) in 100 ml of water is ​

Answer 1

Answer:

Molarity = No of moles of solute ÷ Vol of the solution

Moles of solute that is glucose = mass÷ Molar mass

molar mass of glucose= C6 H12 O6= 12×6 + 1×12 + 16 ×6  = 180g/mol

Moles of glucose= 60÷ 180= 1÷3= 0.333mol

vol of solution=100 ml

1ml =1÷1000l

100ml=1÷1000 × 100

100ml= 0.1 lit

vol of solution= 0.1 litre

Therefore, molarity= 0.333÷0.1

Molarity = 3.33M

Hope it will help

Answer 2

Answer:

Molality solution is 3.33m

Explanation:

Given that,

60gm of glucose is present in 100ml of water

The molecular formula of glucose is

$C _{6}H _{12} O_{6}$

Therefore its molar mass is 180gm

Now,no.of moles of glucose is

$n = \frac{60}{180} = 0.333$

we know that,volume of solvent=100ml

Therefore weight of solvent(water)=100gm

Now we can find molality by using relation

$molality(m) = \frac{no.of \: moles \: of \: solute}{weight \: of \: solvent(in \: gm)} \times 1000$

By substituting given values we get

$m = \frac{0.333}{100} \times 1000 = 3.33$

Therefore molality of solution is 3.33m

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