Molar mass of al(no3)3

What is the mass of Al(NO3)3 9H2O(s) needed to
prepare 50.00 mL of a 0.0650 M solution? (The molar mass of
aluminum nitrate nonahydrate is 375.2 g/mol.) Select one: a. 25.04 g b. 18.76 g c. 1.300 g d. 1.219 g e. 0.375 g

ANS= 1.219g.
First, solve for the moles of Al(NO3)3 dot
50.00 mL x (1 L/1,000 mL) x (0.0650 moles/L) = 0.00325 moles
Next, solve for the mass of Al(NO3)3 dot 9H2O from its
molar mass:
0.00325 moles x (375.2 g/mol) = 1.219 g
The answer is d. 1.219

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