The number of N atoms is 681 g of C 7 H 5 N 3 O 6 is x × 1021. The value of x is ( N A =6.02 ×. 1023 mol -1 ) (Nearest Integer)

Question

The number of N atoms is 681 g of C 7 H 5 N 3 O 6 is x × 1021. The value of x is ( N A =6.02 ×. 1023 mol -1 ) (Nearest Integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

M.M. of C 7 H 5 N 3 O 6 is 84+5+42+96=227 n C 7 H 5 N 3 O 6=(681/227)=3 n N =(681/227) × 3=9 mol no. of N atoms =9 × 6.02 × 1023 =5418 × 1021 ∴ The answer is 5418 .

Q. The number of $N$ atoms is $681 g$ of $C_{7} H_{5} N_{3} O_{6}$ is $x \times 1 0^{21}$ . The value of $x$ is _______ $(N_{A} = 6.02 \times$ $1 0^{23} m o l^{- 1}$ ) (Nearest Integer)

M.M. of $C_{7} H_{5} N_{3} O_{6}$ is $84 + 5 + 42 + 96 = 227$
$n_{C_{7} H_{5} N_{3} O_{6}} = \frac{681}{227} = 3$
$n_{N} = \frac{681}{227} \times 3 = 9 m o l$
no. of $N$ atoms $= 9 \times 6.02 \times 1 0^{23}$ $= 5418 \times 1 0^{21}$
$∴$ The answer is $5418$ .

Q. The number of N atoms is 681g of C7​H5​N3​O6​ is x×1021. The value of x is _______(NA​=6.02× 1023mol−1 ) (Nearest Integer)

M.M. of C7​H5​N3​O6​ is 84+5+42+96=227 nC7​H5​N3​O6​​=227681​=3 nN​=227681​×3=9mol no. of N atoms =9×6.02×1023 =5418×1021 ∴ The answer is 5418 .

Q. The number of $N$ atoms is $681 g$ of $C_{7} H_{5} N_{3} O_{6}$ is $x \times 1 0^{21}$ . The value of $x$ is _______ $(N_{A} = 6.02 \times$ $1 0^{23} m o l^{- 1}$ ) (Nearest Integer)

M.M. of $C_{7} H_{5} N_{3} O_{6}$ is $84 + 5 + 42 + 96 = 227$
$n_{C_{7} H_{5} N_{3} O_{6}} = \frac{681}{227} = 3$
$n_{N} = \frac{681}{227} \times 3 = 9 m o l$
no. of $N$ atoms $= 9 \times 6.02 \times 1 0^{23}$ $= 5418 \times 1 0^{21}$
$∴$ The answer is $5418$ .