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3 years ago

6

A pendulum oscillates 50 times in 6 seconds. Find its time period and frequency?

1 answer:

iren [92.7K]3 years ago

4 0

Explanation:

time taken fir 50 oscillations is 6 seconds

time taken for 1 oscillation is 6/50

convert it into a decimal

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Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o

garri49 [273]

Answer:

$P_{sol}=50.4\ mm.Hg$

Explanation:

According to given:

molecular mass of glycerin, $M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}$

molecular mass of water, $M_w=2+16=18\ g.mol^{-1}$

∵Density of water is $0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}$

∴mass of water in 316 mL, $m_w=316\times 0.992=313.5 g$

mass of glycerin, $m_g=154\ g$

pressure of mixture, $P_x=55.32\ torr= 55.32\ mm.Hg$

temperature of mixture, $T_x=40^{\circ}C$

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

$n_w=\frac{m_w}{M_w}$

$n_w=\frac{313.5}{18}$

$n_w=17.42 moles$

<u>moles of glycerin in the given quantity:</u>

$n_g=\frac{m_g}{M_g}$

$n_g=\frac{154}{92}$

$n_g=1.674 moles$

<u>Now the mole fraction of water:</u>

$X_w=\frac{n_w}{n_w+n_g}$

$X_w=\frac{17.42}{17.42+1.674}$

$X_w=0.912$

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

$\therefore P_{sol}=X_w\times P_x$

$\therefore P_{sol}=0.912\times 55.32$

$P_{sol}=50.4\ mm.Hg$

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