The molecular mass of a compound is 75 (i) Calculate the mass of 100 molecules, in amu. (ii) Calculate the mass of 5000 molecules , in gm. (iii) What

Monjur Alahi

The molecular mass of a compound is 75
(i) Calculate the mass of 100 molecules, in amu.
(ii) Calculate the mass of 5000 molecules , in gm.
(iii) What is the mass of `6.022xx10^(20)` molecules , in gm
(iv) How many molecules are in its 2.5 mg


0 like 0 dislike
19 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

(i) mass of 1 molecules =75 amu
`therefore` mass of 100 molecules=7500 amu
(ii) Mass of 5000 molecules =`5000xx75` amu
`=5000xx75xx1.67xx10^(-24)=6.26 25xx10^(-19)` gm
`because 6.022xx10^(20)` molecules weights `(75)/(6.022xx10^(23))xx6.022xx10^(20)=0.075` gm
(iv) `therefore `75 gm compound contains `6.022xx10^(23)` molecules
`because 2.5xx10^(-3)` gm will contain `(6.022xx10^(23))/(75)xx2.5xx10^(-3)=2.007xx10^(19)` molecules.

0 like 0 dislike
18 views
Talk Doctor Online in Bissoy App
Helium nuclei combines to form an oxygen nucleus. The energy released per nucleon of oxygen nucleus is if `m_(0)=15.834` amu and `m_(He)=4.0026 `amu

Correct Answer - A `4 ._(2)He^(4) rarr _(8)O^(16)` Mass defect `=4 xx 4.0026 -15.834 a.m.u.=0.1764 a.m.u.` Total energy released `=0.1764 xx 931 `MeV Energy released /Nucleon`=(0.1764 xx 931)/(16)=10.24 `MeV

2 Answers 1 views
A nuclear reaction along with the masses of the particle taking part in it is as follows `underset("amu")underset(1.002)A+underset("amu")underset(1.00

Correct Answer - D `Q(1.002+1.004-1.001-1.003)(931.5)MeV=1.863 MeV`

2 Answers 1 views
In a compound C, H and N 1:1 are present in 9 : l : 3.5 by weight. If molecular weight of the compound is 108, the molecular formula of the compound i

Correct Answer - C Ratio of masses :C:H:N =9:1:3.5. Ratio of atoms `:C:H:N=9/12:1/1:3.5/14` `1:1/1xx12/9:3.5/14xx12/9` =1:1.33: 0.33= 3:4:1 E.F. of the compound `=C_3H_4N` E.F mass=`12xx3+4xx1+14xx1` =36+4+14=54 But mol.mass =108 `n=("Mol.mass ")/("E.F. mass...

2 Answers 1 views
An organic compound contains, `C, H` and `S`. The minimum molecular weight of the compound containing `8%` sulphur is : `("atomic weight of" S=32 "amu

Correct Answer - C Minimum molecular mass is obtained when it is supposed that one molecule of the compound contains only one atom of the given element under investigation. `therefore` Min...

2 Answers 1 views
The atomic mass of an element is 50 (i) Calculate the mass of one atom, in amu (ii) Calculate the mass of `6.022xx10^(23)` atoms, in gm (iii) Calculat

(i) 50 amu `" "` (ii) 50 gm (iii) `because` 50 gm of element contains `6.022xx10^(23)` atoms `therefore ` 10 gm of element will contain `(6.022xx10^(23))/(50)xx10=1.2044xx10^(22)` atoms (iv) `because 6.022xx10^(23)`...

2 Answers 1 views
Prove that `^100 C_2^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[^(200)C_(98)-^(100)C_(49)]dot`

To find `S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)` Consider, `.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)` `= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)` = Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)` `= .^(200)C_(98) " "(1)` Also,...

2 Answers 1 views
Find Q1 , D5 and P45 for the following data: Monthly Rent 0-500 500-1000 1000-1500 1500-2000 2000-3000 3000-4000 4000-5000 5000-6000 6000-7000

Q1 = 1862.64, D5 = 2666.67, P45 = 2500

2 Answers 1 views
Calculate the binding energy per nucleon for a `._(6)C^(12)` nucleus. Atomic mass of `._(6)C^(12) = 12am u`, mass of a photon = 1.007825 amu, mass of

Correct Answer - 7.68 MeV

2 Answers 1 views
Calculate the binding energy of a deutron. Given that mass of proton = 1.007825 amu mass of neutron = 1.008665 amu mass of deutron = 2.014103 amu

Correct Answer - 2.2 MeV

2 Answers 1 views
Calculate the Q - value of the reaction `._(1)H^(2)+._(1)H^(2) rarr ._(1)H^(3)+._(1)H^(1)` `m(._(1)H^(2))=2.014103 "amu" m(._(1)H^(3))=3.016049 "amu"

Correct Answer - 4.033 MeV

2 Answers 1 views
X