The water in a hydrate can only be removed by heating the material to very high

Question

The water in a hydrate can only be removed by heating the material to very high temperatures (-1000 degree C). After removal of the water the salt Is referred to as an anhydrate. The general formula for a hydrate is written as: salt-XH_2O where the - indicates that water Is attached to the ions of the salt by weak bonds. A student was asked to identify an unknown hydrate by determining the %H_2O in the hydrate. After heating and cooling, a 2.752 g sample of this unknown weighed 1.941 g. The list of potential unknown was LiNO_3 middot 3H_2O, Ca(NO_3)_2 middot 4H_2O, and Sr(NO_3)_2 middot 4H_2O. Calculate the percent H_2O in the student's unknown. 25.4% 29.5% 32.0% 40.2% What is the probable identity of the students unknown? LiNO_3 middot 3H_2O Ca(NO_3)_2 middot 4H_2O Sr(NO_3)_2 middot 4H_2O None of those listed. The concentration of an acid can be determined by titration with a known concentration of a base. The concentration of a base can be determined by titration with a known concentration of an acid. Neutralization of 0.0015 mole of H_2SO_4 would require 0.15 mole of NaOH. Using phenolphthalein as an indicator of neutralization of a basic solution the end point is the point at which the solution acquires a faint pink color. For a successful titration of an acid by a base the volumes of the acid titrated and the basic solution used in the neutralization must be measured accurately.

Explanation / Answer

Solution:- (7) 1000 degree C is very high and no such higher temperature is required to remove water from a hydrate even when the boiling point of water is only 100 degree C. So, it is false.

(8) There are two thing, hydrate and anhydrate. When the compound has crystaline water in it then we call it hydrate and when this crystaline water is removed then we call it anhydrate. So, it is true.

(9) water is a polar molecule and salts have positive and negative charges so being polar water molecules could gathered around the charged ions. H of water comes closer to the negative end of the salt and O of water comes closer to the positive end of the salt since O has negative charge and H has positive. There could also be like some hydrogen bonding present between the salt and water present in it as crystaline water. These bondings are like weak bonds so the statement is true.

(10) mass of hydrate is given = 2.752 g

mass of sample after heating means the mass of anhydrate = 1.941 g

so, mass of water present in the hydrate is the difference of these two that is, 2.752 g - 1.941 g = 0.811 g

so, percent H2O in the unknown = (mass of water/mass of hydrate) x 100

= (0.811/2.752) x 100 = 29.5%

So, the correct choice is (B) 29.5%.

(11) To get the identity of the unknown from the list of given possible compounds we need to calculate the percentage of water for all of them.

molar mass of LiNO3.3H2O = (6.94 + 14 + 3*16) + 3(2*1.01 + 16) = 68.94 + 54.06 = 123

54.06 g of water are present in 123 g of hydrate. So, percentage of water = (54.06/123) x 100 = 43.95%

molar mass of Ca(NO3)2.4H2O = (40.08 + 28 + 96) + 4(18.02) = 164.08 +72.08 = 236.16

water percentage = (72.08/236.16) x 100 = 30.5%

molar mass of Sr(NO3)2.6H2O = (87.62 + 28 + 96) + 6(18.02) = 211.62 + 108.12 = 319.74

water percentage = (108.12/319.74) x 100 = 33.8%

experimental value for water percentage (29.5%) is closer to the water percentage(30.5% in Ca(NO3)2.4H2O). So, the propable indentity of the students unknown is Ca(NO3)2.4H2O.

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