How many grams of the molecule glucose C6H12O6 would be required to make 1 L of a 0.5 M solution of the molecule?

90g

Explanation:

We need to find the molar mass of C6H12O6. This is done below:

MM of C6H12O6 = (12x6) + (12x1) + (6x16) = 72 + 12 + 96 = 180g/mol

From the question, we obtained:

Volume = 1L

Molarity = 0.5M

Number of mole = Molarity x Volume

Number of mole = 0.5 x 1 = 0.5mol

Mass conc of C6H12O6 = number of mole x MM = 0.5 x 180 = 90g

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.

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