Question

How many gram of solute is required to prepare 1.0 L of 1 M CaCl2.6H2O? *

Answer 1

Answer:

Answer:

molar mass ofCaCl2.6H2O = 40 + 2(35.5) + 6(18)

                                              = 219 g mol^-1

Molarity = weight of solute * 1000 / molar mass of solute * volume

  1 = weight of solute * 1 / 219 * 1

weight of solute = 219 gm

here volume is already given in litre so we don't need to multiply with (1000).

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Answer 2

Answer:

Here,

s=1M

v=1.0L

m=219

We know,

weight of solute,w=svm=1*1*219=219g(Ans)

Explanation:

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