Question

13. What is the molality of a solution with 18.4 moles LiF in 26 kg H2O? *

Answer 1

Here

• Solute=Lithium fluoride=LiF
• Solvent=Water=H_2O

• Moles of Solute=18.4mol
• Volume of water=26kg=26dm^3 or 26L(1kg of water weighs 1dm^3 at max density i.e at -4°C)

Now

$\boxed{\mathcal{ Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\: solvent\:in\:L}}}$

Put the values

$\\ \rm\hookrightarrow Molarity=\dfrac{18.4}{26}$

$\\ \rm\hookrightarrow Molarity=0.707M$

$\\ \rm\hookrightarrow Molarity=0.71M$

Option A is correct

Answer 2

Here, We are asked to calculate the molarity of a solution with 18.4 moles of Lithium Fluoride in 26 kg of water... To find the molarity, We need to divide the number of moles by the volume of solution in liters

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• Number of moles = 18.4
• Volume of solution = 26 kg = 37.14 liters

$\tt \twoheadrightarrow \: molarity = \frac{number \: of \: moles}{volume \: of \: solution}$

$\tt \twoheadrightarrow\: molarity = \frac{18.4}{37.14}$

$\sf \Rrightarrow \: molarity = 0.71 \: molar$

➪ Therefore, The molarity of solution is 0.71 molar...~