few gchem Q's

[utdent20]

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If 14 grams of NaF are dissolved in 400 milligrams of water , what is the resulting molality of the solutions? a) .4, b).83, c).93, D).54, E).32

2. How much 4M Ca(OH)2 is needed to neutralize 300 milliliteres of3M HNO3?
a. 234 b. 655, c. 623. d.723, e. 112.5

On the first one, I tried to put it in the formula and got
14g NaF X 1 mol/ 32 g leading to 7/(16X.400) which did not give any of the answers..😡
second one, using the M1V1=M2V2 did not work, so then using normality should help? but i cannot figure how many times to use the H as in twice due to OH2 or thrice coz of total H's being used in the problem..

 

[211183]

1. molar mass of Na F is not 32...

2. i think its 300* 3M / 4M*2

someone can confirm my #2

 

[DDS13]

exactly, u r right on both of them

 

[Contach]

If 14 grams of NaF are dissolved in 400 milligrams of water , what is the resulting molality of the solutions? a) .4, b).83, c).93, D).54, E).32

2. How much 4M Ca(OH)2 is needed to neutralize 300 milliliteres of3M HNO3?
a. 234 b. 655, c. 623. d.723, e. 112.5

1. molar mass of NaF is 23 + 19 = 42 g/mol
mols = 14g / 42g/mol = 1/3 mol
molality = mols solute / kg solvent
400 miligrams of water = 400 milligrams x 1 gram/1000milligrams x 1kg/1000gram = 400 x 10^-6 = 4 x 10^-4 kg
so molality = 1/3 mol / 4 x 10^-4 kg = 1/3 x 1/0.0004 = 0.0012 m = none of the answers.. haha..

2. N1V1 = N2V2
4M*2(V1) = 3M(0.3)
V1 = 0.9/8 about 1/8 is about .125.. maybe a bit lower.. that thats in L.. looks like the answers are in mL.. so e. 112.5

 

[211183]

if its 400 g, you get .83. a typo?

 

[Contach]

if its 400 g, you get .83. a typo?

oh true, i forgot the 1/ 0.00012 = 0.83 on my CALCULATOR. how would one do this without a calculator???

 

[211183]

don't know, i used one.

when using R

do you simplify .0821
to

8.2 *10^-2 is that hte easiest way to do that math?

 

[Contach]

don't know, i used one.

when using R

do you simplify .0821
to

8.2 *10^-2 is that hte easiest way to do that math?

i try very hard not to use R.. you usually can get around it by doing the STP numbers.. because r is just a ratio of PV/T at STP.. try it out... but if i have to i simplify to 0.08 haha...

 

[utdent20]

thanks soo much on your input guys... it was hard not to use a calculator on these.. they are out of barrons's but.. hopefully the real thing would not give us such hard calculations.

 

[utdent20]

the answer to #1 is B. .83.. I dont know hwo to get that?

 

[joonkimdds]

so there is no answer for the 1st one??

 

[211183]

its .83 my best guess, i wouldn't trust barrons as i was going through it sa wa lot of errors. also ahve heard the same from others.

 

[joonkimdds]

but it can't be 0.83 becuz it says miligram, not gram.

 

[211183]

i know i said if it was grams it works out to . 83. i have no explanation besides that it is a typo. and i came across a lot of them in barrons.

I can't be sure though I guess.....unless someone has another way to figure it uot.
😕