G.com what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume
1 answer:
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Answer:
i. Molar mass of glucose = 180 g/mol
ii. Amount of glucose = 0.5 mole
Explanation:
<em>The volume of the glucose solution to be prepared</em> = 500
<em>Molarity of the glucose solution to be prepared</em> = 1 M
i. Molar mass of glucose () = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
ii.<em> mole = molarity x volume</em>. Hence;
amount (in moles) of the glucose solution to be prepared
= 1 x 500/1000 = 0.5 mole