Number of moles of solute present in $1000$ $g m$ of solvent is known as :

Normality

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Molality

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Molarity

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Mole fraction

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Solution

The correct option is C Molality
The number of moles of solute present in 1000 gm (or 1 kg) of solvent is known as molality.
$Molality = \frac{Number of moles of solute}{Mass of solvent (in kg)}$
$Molarity = \frac{Number of moles of solute}{volume of solution (in L)}$
$Normality = \frac{Number of gram equivalents of solute}{volume of solution (in L)}$
$Mole fraction = Number of moles of solute Number of moles of solute + Number of moles of solvent$

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Similar questions

\frac{N u m b e r o f m o l e s o f s o l u t e}{N u m b e r o f k i l o g r a m s o f t h e s o l v e n t}

w_{A}

m_{A}

w_{B}

\frac{w_{A}}{m_{A} \times w_{B}} \times 1000

X_{A} = \frac{n}{N + n} a n d X_{B} = \frac{N}{N + n}

\frac{X_{A}}{X_{B}} = \frac{n}{N} = \frac{M o l e s o f s o l u t e}{M o l e s o f s o l v e n t} = \frac{W_{A} \times m_{B}}{w_{B} \times m_{A}}

\frac{X_{A} \times 1000}{X_{B} \times m_{B}} = \frac{w_{A} \times 1000}{W_{B} \times m_{A}} = m

\frac{X_{A} \times 1000}{(1 - X_{A}) m_{B}} = m

\frac{1}{3}

\frac{1}{2}

800

A solution A has 10 gm of solute B dissolved in 2 moles of solvent. What is the mole fraction of the solute?

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