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marissa [1.9K]
3 years ago
10

Which of the following statements best describes why a solution of 6.00 g of Ca(NO3)2?in 30.0 g of water has a greater boiling-p

oint elevation than a solution of 3.00 g of Ca(NO3)2?in 60.0 g of water?
Chemistry
1 answer:
sukhopar [10]3 years ago
5 0
Boiling-point elevation is a colligative property.

That means, the the boiling-point elevation depends on the molar content (fraction) of solute.

The dependency is ΔTb = Kb*m

Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.

A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:

(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg =   50/molar mass

=> 200 / 50 = 4.

Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g  of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.

Or, what is the same, the solution with higher molality will have the higher boiling point.
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In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat
LiRa [457]

Answer: 1.14

Explanation:

HCl+NaOH\rightarrow NaCl+H_2O

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL

Putting values in above equation, we get:

1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M

To calculate pH of gastric juice:

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3 0
3 years ago
How many grams of Pb are contained in a mixture of 0.135 kg each of PbCl(OH) and Pb2Cl2CO3?
Dahasolnce [82]

Hey there!:

Given the mass of PbCl(OH) :

0.135 Kg = 0.135 Kg*(1000g / 1Kg)  = 135 g

Molecular mass of PbCl(OH)  = 207+35.5+16+1  = 259.5 g / mol

Atomic mass of Pb = 207 g/mol

Hence mass of Pb in 135 g  PbCl(OH)  :

(207 g Pb /  259.5 g PbClOH) * 135g PbClOH  =

0.79768 * 135 =>  107.68 g of Pb

For Pb2Cl2CO3  :

Given the mass of Pb2Cl2CO3  :

0.135 Kg = 0.135 Kgx(1000g / 1Kg)  = 135 g

Molecular mass of Pb2Cl2CO3  = 2*207+2*35.5+12+3*16  = 545 g / mol

Mass of Pb present in 1 mol (=545 g / mol) of Pb2Cl2CO3  = 2*207 = 414 g

Hence mass of Pb in 135 g  Pb2Cl2CO3:

(414 g Pb /  545 g PbClOH) * 135g PbClOH  =

0.75963 * 135 => 102.55 g of Pb2Cl2CO3


Hope that helps!

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