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When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier than ordinary seawater) is evaporated, 196 grams of MgCl2 are recovered. What is the molarity of MgCl2 in the original sample?

Answer :

Answer:

2.06 M

Explanation:

number of moles of MgCl₂ = mass given in gram / molar mass

molar mass of MgCl₂ = 95.211 g/mol

number of moles of MgCl₂ = 196 g / 95.211 g/mol = 2.06 mol

Molarity of MgCl₂ in the original sample = number of moles / volume in Liters = 2.06 mol / 1 L = 2.06 M

Molarity represent the number of solutes per liter of solution. The molarity of the solution is 2.06 M.

Molarity:

It can be understood as how many moles of solute present in the 1 liter of solution. It is represented by the M.

The molarity of the formula is,

[tex]\bold{ M = \frac{W \times 1000}{M \times V} }[/tex]

Where,

W- number of moles given mass

M- molar mass  [tex]\bold{ MgCl_2}[/tex] =  95.211 g/mol

V-  given volume

Put the values in the formula.

[tex]\bold {M = \frac{196\times 1000}{95.211 \times 1000} }\\\\\bold {M = 2.06 }[/tex]

Therefore the molarity of the solution is 2.06 M.

To know more about Molarity, refer the link:

https://brainly.com/question/8732513

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