$171 g$ of cane sugar $(C_{12} H_{22} O_{11})$ is dissolved in $1$ litre of water. The molarity of the solution is.

2.0 M

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1.0 M

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0.5 M

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0.25 M

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Solution

The correct option is C $0.5 M$
$Molarity is defined as the number of moles of solute present in a liter of solution.$

$M a t h e m a t i c a l l y,$

$M o l a r i t y = \frac{No. of mole of solute}{Volume of solution (L)}$

Hence, $M o l a r i t y = \frac{171 / 342}{1}$

$= 0.5$

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171 g of cane sugar (C12H22O11) is dissolved in 1 litre of water. The molarity of the solution is.

The correct option is C 0.5 MMolarity is defined as the number of moles of solute present in a liter of solution.Mathematically, Molarity=No. of mole of soluteVolume of solution(L) Hence, Molarity=171/3421 =0.5

$171 g$ of cane sugar $(C_{12} H_{22} O_{11})$ is dissolved in $1$ litre of water. The molarity of the solution is.

The correct option is C $0.5 M$
$Molarity is defined as the number of moles of solute present in a liter of solution.$

$M a t h e m a t i c a l l y,$

$M o l a r i t y = \frac{No. of mole of solute}{Volume of solution (L)}$

Hence, $M o l a r i t y = \frac{171 / 342}{1}$

$= 0.5$