If 36.0 g of glucose is present in 400 ml of solution, molarity of the solution is:

0.05 M

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11.0 M

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0.5 M

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2.0 M

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Solution

The correct option is D $0.5 M$
Molecular weight of glucose = 180 gm
Given,
Mass of solute = 36 g
Volume of solution = 400 ml = 0.4 L

Molarity = $\frac{No of moles}{Volume(in L)}$

$= \frac{36}{180 \times 0.4}$

$= 0.5 M$

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If 36.0 g of glucose is present in 400 ml of solution, molarity of the solution is:

The correct option is D 0.5MMolecular weight of glucose = 180 gmGiven,Mass of solute = 36 gVolume of solution = 400 ml = 0.4 LMolarity = No of molesVolume(in L) =36180×0.4 =0.5M

The correct option is D $0.5 M$
Molecular weight of glucose = 180 gm
Given,
Mass of solute = 36 g
Volume of solution = 400 ml = 0.4 L

Molarity = $\frac{No of moles}{Volume(in L)}$

$= \frac{36}{180 \times 0.4}$

$= 0.5 M$