Question

can u pls solve these numericals i tried i dudnt get these at ideal gas law ones pv = nrt no irrelevant answers ​

Answer 1

Answer:

1. Given: weight of gas 280ml at 305k and 750mm of Hg is 0.344

Find: molecular weight of the gas.

solution:

According to Ideal gas equations

$pv = \frac{w}{m} rt \\ p = \frac{750}{760} = 0.987atm \\ v = \frac{280}{1000} = 0.28l \\ we \: know \: that n = \frac{w}{m} \frac{mass \: of \: gas}{molecular \: mass } \\ w = 0.344g \\ r = gas \: constant \: = 0.0821l \: atm \: k \: mol \\ t \: = absolute \: temperature \: = 305k \\ substitute. \\ 0.987 \times 0.28 = \frac{0.344}{m} (0.0821 \times 305) \\ m = (1.249)(25.04) \\ m = 21.19 \\ molecular \: weight = 21.19$

T= Absolute temperature=305k

substitute:

$0.987 \times 0.28 = \frac{0.344}{m} (0.0821 \times 305) \\ m = 1.246)(25.04) \\ m = 21.19$

2. P=16atm

V=9.00l

T=27°C=300k

w=93.6g

According to the gas equations,

$pv = \frac{w}{m}rt \\ or \: m = \frac{wrt}{pv} \\ substituting \: the \: value \: we \: get \: \\ m = \frac{93.6 \times 0.0821 \times 300}{16 \times 9.00} \: = 16.01$

molecular mass=16.01

Explanation:

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