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3 years ago

You do an experiment in which you need 0.5 moles of tyrosine (C9H11NO3). How many grams must you weigh out?

Chemistry

1 answer:

Yuki888 [10]3 years ago

5 0

Answer: a) 90.5g

b) 33.6 L

Explanation:-

Molar mass of tyrosine $(C_9H_{11}NO_3)$ = 181 g/mol

According to Avogadro's law, 1 mole of every substance weighs equal to its molar mass.

1 mole of tyrosine $(C_9H_{11}NO_3)$ weighs = 181 g/mol

0.5 moles of tyrosine $(C_9H_{11}NO_3)$ weigh $=\frac{181g/mol}{1}\times 0.5moles=90.5g$

b) According to Avogadro's law, 1 mole of an ideal gas occupies 22.4 Liters at Standard conditions of temperature and pressure (STP).

1 mole of gas at STP occupy = 22.4 L

1.5 moles of gas at STP occupy = $\frac{22.4L}{1}\times 1.5=33.6L$

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