QUES NO.-2327,
How many gram of solute is required to prepare 600 mL of 1.15 M CaCl2.6H20?
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Answer:
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151.1 gram of solute is required to prepare 600ml of solution.
Step-by-step process
Given:
Molarity of solution= 1.15M
Volume of solution= 600ml
To find= the grams of solute
The molar mass of the solute is given by= 40+ 2(35.5)+ 6(18)
=219 gram/mol
Solution:
The formula for molarity:
Molarity= weight of solute × 1000÷ molar mass of solute × volume
Thus, plugging in the values in the above-mentioned formula
1.15= weight of solute × 1000 ÷ 219 × 600
1.15 × 131.4 = weight of solute
Result:
151.1 gram= weight of solute
151.1 gram of solute is required to prepare 600ml of solution.
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