What mass of solute exists in a $1L$ volume of $0.650mol*L^-1$ $CaBr_2$ ?

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Answer 1 · 2017-01-05T15:48:38

Approx. $130*g$ .

$"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$ , and, clearly, this has units of $mol*L^-1$ .

Here we have a $1L$ volume of $0.650*mol*L^-1$ $CaBr_2(aq)$ .

The product,

$"Volume of solution "xx" Concentration"$ $=$ $"Moles of solute"$

And thus, $1*cancelLxx0.650*mol*cancel(L^-1)=0.650*mol$ $CaBr_2$ .

And $0.650*cancel(mol)xx199.89*g*cancel(mol^-1)=??*g$

What mass of solute exists in a 1*L1*L volume of 0.650*mol*L^-10.650*mol*L^-1 CaBr_2CaBr_2?