Time required : 3 s
<h3>Further explanation
</h3>
Power is the work done/second.
![\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s](https://tex.z-dn.net/?f=%5Ctt%20P%3D%5Cdfrac%7BW%7D%7Bt%7D%5C%5C%5C%5CP%3Dpower%2Cj%2Fs%2Cwatt%5C%5C%5C%5CW%3Dwork%2C%20J%5C%5C%5C%5Ct%3Dtimes%3Ds)
To do 33 J of work with 11 W of power
P = 11 W
W = 33 J
![\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}](https://tex.z-dn.net/?f=%5Ctt%20t%3D%5Cdfrac%7BW%7D%7BP%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B33%7D%7B11%7D%5C%5C%5C%5Ct%3D%5Cboxed%7B%5Cbold%7B3~s%7D%7D)
Answer:
The answer to your question is: V2 = 1 l
Explanation:
Data
P1 = 200 kPa
P2 = 300 kPa
V1 = 1.5 l
V2 = ?
Formula
P1V1 = P2V2
V2 = (P1V1) / P2
V2 = (200 x 1.5) / 300
V2 = 1 l
Answer:
![P_{sol}=50.4\ mm.Hg](https://tex.z-dn.net/?f=P_%7Bsol%7D%3D50.4%5C%20mm.Hg)
Explanation:
According to given:
- molecular mass of glycerin,
![M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}](https://tex.z-dn.net/?f=M_g%3D3%5Ctimes%2012%2B8%2B3%5Ctimes%2016%3D92%5C%20g.mol%5E%7B-1%7D)
- molecular mass of water,
![M_w=2+16=18\ g.mol^{-1}](https://tex.z-dn.net/?f=M_w%3D2%2B16%3D18%5C%20g.mol%5E%7B-1%7D)
- ∵Density of water is
![0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}](https://tex.z-dn.net/?f=0.992%5C%20g.cm%5E%7B-3%7D%3D%200.992%5C%20g.mL%5E%7B-1%7D)
- ∴mass of water in 316 mL,
![m_w=316\times 0.992=313.5 g](https://tex.z-dn.net/?f=m_w%3D316%5Ctimes%200.992%3D313.5%20g)
- mass of glycerin,
![m_g=154\ g](https://tex.z-dn.net/?f=m_g%3D154%5C%20g)
- pressure of mixture,
![P_x=55.32\ torr= 55.32\ mm.Hg](https://tex.z-dn.net/?f=P_x%3D55.32%5C%20torr%3D%2055.32%5C%20mm.Hg)
- temperature of mixture,
![T_x=40^{\circ}C](https://tex.z-dn.net/?f=T_x%3D40%5E%7B%5Ccirc%7DC)
<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}](https://tex.z-dn.net/?f=n_w%3D%5Cfrac%7Bm_w%7D%7BM_w%7D)
![n_w=\frac{313.5}{18}](https://tex.z-dn.net/?f=n_w%3D%5Cfrac%7B313.5%7D%7B18%7D)
![n_w=17.42 moles](https://tex.z-dn.net/?f=n_w%3D17.42%20moles)
<u>moles of glycerin in the given quantity:</u>
![n_g=\frac{m_g}{M_g}](https://tex.z-dn.net/?f=n_g%3D%5Cfrac%7Bm_g%7D%7BM_g%7D)
![n_g=\frac{154}{92}](https://tex.z-dn.net/?f=n_g%3D%5Cfrac%7B154%7D%7B92%7D)
![n_g=1.674 moles](https://tex.z-dn.net/?f=n_g%3D1.674%20moles)
<u>Now the mole fraction of water:</u>
![X_w=\frac{n_w}{n_w+n_g}](https://tex.z-dn.net/?f=X_w%3D%5Cfrac%7Bn_w%7D%7Bn_w%2Bn_g%7D)
![X_w=\frac{17.42}{17.42+1.674}](https://tex.z-dn.net/?f=X_w%3D%5Cfrac%7B17.42%7D%7B17.42%2B1.674%7D)
![X_w=0.912](https://tex.z-dn.net/?f=X_w%3D0.912)
<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](https://tex.z-dn.net/?f=%5Ctherefore%20P_%7Bsol%7D%3DX_w%5Ctimes%20P_x)
![\therefore P_{sol}=0.912\times 55.32](https://tex.z-dn.net/?f=%5Ctherefore%20P_%7Bsol%7D%3D0.912%5Ctimes%2055.32)
![P_{sol}=50.4\ mm.Hg](https://tex.z-dn.net/?f=P_%7Bsol%7D%3D50.4%5C%20mm.Hg)
4. E
5. D
6. F
Hope this helps